A large object with an initial temperature of $\mathbf{1 5 0}$ degrees Fahrenheit is dropped into the ocean. The current temperature of the ocean water in the area is 77 degrees. The function $f(t)=C e^{(-k t)}+77$ represents the situation, where $t$ is time in minutes, $C$ is a constant, and $k$ is a constant.

After $\mathbf{1 0}$ minutes the object has a temperature of $\mathbf{1 2 0}$ degrees. After how many minutes will the temperature of the object be equal to 80 degrees? Round your answer to the nearest whole number, and do not include units.

Provide your answer below:

Step 1 ：First, we need to find the values of the constants C and k. We know that at t=0, the temperature of the object is 150 degrees. So we can substitute these values into the equation: \(150 = Ce^0 + 77\)

Step 2 ：Solving for C, we get \(C = 150 - 77 = 73\)

Step 3 ：Next, we know that after 10 minutes, the temperature of the object is 120 degrees. So we can substitute these values into the equation: \(120 = 73e^{-10k} + 77\)

Step 4 ：Solving for e^{-10k}, we get \(e^{-10k} = \frac{43}{73}\)

Step 5 ：Taking the natural logarithm of both sides, we get \(-10k = \ln\left(\frac{43}{73}\right)\)

Step 6 ：Solving for k, we get \(k = -\frac{\ln\left(\frac{43}{73}\right)}{10}\)

Step 7 ：Now that we have the values of C and k, we can find the time it takes for the temperature of the object to reach 80 degrees. We set the function equal to 80 and solve for t: \(80 = 73e^{-kt} + 77\)

Step 8 ：Solving for e^{-kt}, we get \(e^{-kt} = \frac{3}{73}\)

Step 9 ：Taking the natural logarithm of both sides, we get \(-kt = \ln\left(\frac{3}{73}\right)\)

Step 10 ：Solving for t, we get \(t = -\frac{\ln\left(\frac{3}{73}\right)}{k}\)

Step 11 ：Substituting the value of k we found earlier, we get \(t = 10 \frac{\ln\left(\frac{3}{73}\right)}{\ln\left(\frac{43}{73}\right)}\)

Step 12 ：Using a calculator, we find that t is approximately 47. Therefore, it will take about 47 minutes for the temperature of the object to reach 80 degrees. So, \(\boxed{t \approx 47}\)